TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Complete integrals and covariant versus time Hamilton-Jacobi

Statement

Assume ${A C}_{ω}$ for the cotangent/HJ supplier chain and let $n \geq 1$ . Let $U_{H} \subset R \times T^{*} R^{n}$ be open and $H \in C^{\infty} (U_{H}, R)$ . Let $D_{S} \subset R \times R_{q}^{n} \times R_{α}^{n}$ be open and $S \in C^{3} (D_{S}, R)$ , with $S_{t} + H (t, q, S_{q}) = 0$ wherever $(t, q, α) \in D_{S}$ and $(t, q, S_{q} (t, q, α)) \in U_{H}$ . If the mixed matrix $S_{q α}$ is invertible at $(t, q, α)$ , then after restricting neighborhoods the equations $p_{i} = S_{q^{i}}$ , $β^{i} = S_{α_{i}}$ define a local $C^{2}$ diffeomorphism $F_{t} : (q, p) \mapsto (β, α)$ at each fixed $t$ . Equip the old chart $(q, p)$ with $ω_{o l d} = \sum_{i} d q^{i} \land d p_{i}$ and the target chart, whose position coordinates are $β^{i}$ and momentum coordinates are $α_{i}$ , with $ω_{n e w} = \sum_{i} d β^{i} \land d α_{i}$ . Then $F_{t}^{*} ω_{n e w} = ω_{o l d}$ , so this is a local canonical transformation in the stated convention, and $β$ is constant along the Hamilton characteristics $\dot{q} = H_{p}$ , $\dot{p} = - H_{q}$ . Such a family is a complete integral only on that nonsingular local domain. In the physical interpretation, $q^{i}, β^{i}$ have units m, $p_{i}, α_{i}$ have units kg m/s, $t$ has units s, and $S$ has units J s. $S$ is assigned action units in this interpretation.

For the relativistic specialization, take $n = 3$ , $m \geq 0$ in kilograms and $c > 0$ in metres per second, and a temporal product chart $(x^{0}, q) \in I_{0} \times Ω$ with the selected future normal of A curved temporal chart gives a local reduced Hamiltonian. Write its metric as $g = - N^{2} (d x^{0})^{2} + h_{i j} (d q^{i} + b^{i} d x^{0}) (d q^{j} + b^{j} d x^{0})$ , where $N > 0$ , $b^{i}$ and the positive-definite symmetric matrix $h_{i j}$ are smooth; $h^{i j}$ is the inverse matrix. The notation $b^{i}$ denotes the shift, distinct from the integration coordinate $β^{i}$ . Set $x^{0} = c t$ and require the Hamiltonian in this specialization to be exactly

$H (t, q, p) = c (N (c t, q) \sqrt{m^{2} c^{2} + h^{i j} (c t, q) p_{i} p_{j}} - b^{i} (c t, q) p_{i}),$

on $U_{H} = {t : c t \in I_{0}} \times T^{*} Ω$ for $m > 0$ , or its open part $p \neq 0$ for $m = 0$ . For each fixed $α$ , define the spacetime action $S (x^{0}, q, α) : = S (x^{0} / c, q, α)$ on the corresponding chart domain and write $S_{a} : = \partial_{x^{a}} S$ . Then the covariant equation $g^{a b} S_{a} S_{b} + m^{2} c^{2} = 0$ with $(d S)^{♯}$ future, excluding the zero covector when $m = 0$ , is equivalent to $S_{t} + H (t, q, S_{q}) = 0$ , with $p_{0} = S_{t} / c = - H / c$ . The canonical-coordinate result above still applies to arbitrary $H$ ; the covariant-shell equivalence uses this specified reduced Hamiltonian.

Facts & Assumptions

Given: $n \geq 1$ , the open domains and smooth Hamiltonian in the statement, a local $C^{3}$ HJ solution, and a point where $S_{q α}$ is invertible.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): assume countable choice ${A C}_{ω}$ only for the cotangent and HJ supplier chain; no full axiom of choice is used.

[L1]

Local Hamilton-Jacobi solution by invertible characteristics: for each initial point $(0, a_{0}, d S_{0} (a_{0}))$ lying in its open Hamiltonian domain, the local characteristic construction supplies an actual smooth HJ solution while its projection remains invertible; no continuation beyond loss of projection invertibility is asserted.

[F1]

The Euclidean inverse function theorem: a $C^{1}$ map with invertible derivative has a local $C^{1}$ inverse $g$ , with $D g (y) = D f (g (y))^{- 1}$ . The scalar derivative algebra and total chain rule are supplied by Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ and The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ .

[L2]

A curved temporal chart gives a local reduced Hamiltonian: the current H9 draft assumes a temporal product chart, $N > 0$ , positive spatial metric, and excludes $p_{i} = 0$ on the massless branch; it gives $p_{0} = β^{i} p_{i} - N \sqrt{m^{2} c^{2} + h^{i j} p_{i} p_{j}}$ and $H = - c p_{0}$ . The supplier is a mathematical deduction within its stated chart, positive-root and nonzero-null hypotheses; it introduces no physical action adoption. Here the temporal coefficients in the reduced root are evaluated at $(c t, q)$ ; the physical-time domain is ${t : c t \in I}$ .

Proof

technique · implicit-function and generating-function identities

Given: The hypotheses in the statement and exact conditional temporal-shell interface [L2].

1.1F1givenalgebra

First obtain the exact finite-regularity inverse used here from [F1]. If $f$ is $C^{2}$ , its local $C^{1}$ inverse $g$ satisfies $D g = (D f \circ g)^{- 1}$ . The entries of $D f \circ g$ are $C^{1}$ by the chain rule. Matrix inversion on invertible matrices is $C^{1}$ entrywise by the cofactor-over-determinant formula, whose denominator is nonzero; finite products and reciprocal differentiation use the derivative algebra in [F1]. Consequently $D g$ is $C^{1}$ , and $g$ is $C^{2}$ . At fixed $t$ , the map $(q, α) \mapsto (q, p = S_{q})$ has derivative $(\begin{array}{cc} I & 0 \\ S_{q q} & S_{q α} \end{array})$ , which is invertible at the stated point. By [F1], after restriction it is a diffeomorphism and solves locally for $α = α (t, q, p)$ . Keeping $t$ as an additional identity coordinate in $(t, q, α) \mapsto (t, q, S_{q})$ gives the same invertible block derivative and shows that the inverse $α (t, q, p)$ is jointly $C^{2}$ . Define $β = S_{α} (t, q, α (t, q, p))$ ; this gives a $C^{2}$ local map $(q, p) \mapsto (β, α)$ .

1.2L2A1givenalgebra

In the relativistic specialization, the chain rule gives $p_{0} = \partial_{x^{0}} S = S_{t} / c$ and $p_{i} = \partial_{q^{i}} S = S_{q^{i}}$ . The future-shell root [L2] is $p_{0} = b^{i} (c t, q) p_{i} - N (c t, q) \sqrt{m^{2} c^{2} + h^{i j} (c t, q) p_{i} p_{j}} = - H (t, q, p) / c$ for the Hamiltonian explicitly required in the statement. Thus the covariant future-shell equation implies $S_{t} + H (t, q, S_{q}) = 0$ . Conversely, this HJ equation sets $p_{0}$ to precisely that root, which solves the covariant shell equation and selects its future branch; the root satisfies $p (n) = - \sqrt{m^{2} c^{2} + h^{i j} p_{i} p_{j}} < 0$ for the selected future unit normal $n$ . The nonzero-momentum domain when $m = 0$ excludes the zero null covector.

2.1F1step 1.1algebra

The map $(q, α) \mapsto (β = S_{α}, α)$ has derivative $(\begin{array}{cc} S_{α q} & S_{α α} \\ 0 & I \end{array})$ , which is invertible because $S_{α q} = (S_{q α})^{T}$ . By [F1] it has a $C^{2}$ local inverse solving for $q$ from $(β, α)$ ; then $p = S_{q}$ gives the $C^{2}$ inverse map to the one in step 1.1. The equations therefore define the claimed local diffeomorphism.

2.2L1A1step 1.1algebra

Along a Hamilton characteristic, $\dot{q} = H_{p}$ and $\dot{p} = - H_{q}$ . Differentiate $p = S_{q} (t, q, α (t, q, p))$ along it, and differentiate the HJ equation in $q$ to obtain $\dot{p} = - H_{q} + S_{q α} \dot{α}$ . Since $S_{q α}$ is invertible, $\dot{α} = 0$ . Differentiate the HJ equation with respect to $α$ : $S_{α t} + S_{α q} H_{p} = 0$ . Thus $\dot{β} = \frac{d}{d t} S_{α} = S_{α t} + S_{α q} \dot{q} = 0$ .

3.1givenstep 1.1step 2.1algebra

At fixed $t$ , the differential in $(q, α)$ is $d_{(q, α)} S = p_{i} d q^{i} + β^{i} d α_{i}$ . Taking its exterior derivative gives $0 = d p_{i} \land d q^{i} + d β^{i} \land d α_{i}$ , hence $\sum_{i} d q^{i} \land d p_{i} = \sum_{i} d β^{i} \land d α_{i}$ . This is exactly $F_{t}^{*} ω_{n e w} = ω_{o l d}$ for the source and target forms stated above.

4.1givenalgebra∎

Under the stated SI assignments, $S$ has joule-seconds, so $p = S_{q}$ has kg m/s and $S_{t}$ has joules. Since $α$ is assigned kg m/s, $β = S_{α}$ has metres. Thus the HJ equation and the canonical generating identity respect the assigned action units.

Complete integrals and covariant versus time Hamilton-Jacobi

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources